Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 14 Chemical Kinetics In kinetics we study the rate at which a chemical process occurs. Lecture Presentation © 2012 Pearson Education, Inc.

Similar presentations


Presentation on theme: "Chapter 14 Chemical Kinetics In kinetics we study the rate at which a chemical process occurs. Lecture Presentation © 2012 Pearson Education, Inc."— Presentation transcript:

1 Chapter 14 Chemical Kinetics In kinetics we study the rate at which a chemical process occurs. Lecture Presentation © 2012 Pearson Education, Inc.

2 Chemical Kinetics Factors That Affect Reaction Rates 1.Physical state of the reactants. –In order to react, molecules must come in contact with each other. –The more homogeneous the mixture of reactants, the faster the molecules can react. © 2012 Pearson Education, Inc.

3 Chemical Kinetics Factors That Affect Reaction Rates 2. Concentration of reactants. –As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. © 2012 Pearson Education, Inc.

4 Chemical Kinetics Factors That Affect Reaction Rates 3. Temperature –At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. © 2012 Pearson Education, Inc.

5 Chemical Kinetics Factors That Affect Reaction Rates 4. Presence of a catalyst. –Catalysts speed up reactions by changing the mechanism of the reaction. –Catalysts are not consumed during the course of the reaction. © 2012 Pearson Education, Inc.

6 Chemical Kinetics Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative. 13.1

7 Chemical Kinetics Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate =  1a1a  [A]  t =  1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t = © 2012 Pearson Education, Inc.

8 Chemical Kinetics Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate =  [C 4 H 9 Cl]  t

9 Chemical Kinetics Reaction Rates Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) © 2012 Pearson Education, Inc.

10 Chemical Kinetics Reaction Rates A plot of [C 4 H 9 Cl] versus time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) © 2012 Pearson Education, Inc.

11 Chemical Kinetics Reaction Rates All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) © 2012 Pearson Education, Inc.

12 Chemical Kinetics Reaction Rates In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Rate =  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t © 2012 Pearson Education, Inc.

13 Chemical Kinetics The Rate Law 13.2 The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers (determined experimentally). aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall

14 Chemical Kinetics F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1 13.2

15 Chemical Kinetics Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2  (aq) N 2 (g) + 2 H 2 O(l) © 2012 Pearson Education, Inc.

16 Chemical Kinetics Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2  ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2  (aq) N 2 (g) + 2 H 2 O(l) © 2012 Pearson Education, Inc.

17 Chemical Kinetics Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2  ] Rate  [NH 4 + ] [NO 2  ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2  ] This equation is called the rate law, and k is the rate constant (depends only on reagents). Therefore, © 2012 Pearson Education, Inc.

18 Chemical Kinetics Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k[NH 4 + ] [NO 2  ] the reaction is First-order in [NH 4 + ] and First-order in [NO 2  ] © 2012 Pearson Education, Inc.

19 Chemical Kinetics Rate Laws Rate = k[NH 4 + ] [NO 2  ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall. © 2012 Pearson Education, Inc.

20 Chemical Kinetics F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ] 13.2

21 Chemical Kinetics Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 13.3

22 Chemical Kinetics Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0. © 2012 Pearson Education, Inc.

23 Chemical Kinetics Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 =  kt where [A] 0 is the initial concentration of A, and [A] t is the concentration of A at some time, t, during the course of the reaction. © 2012 Pearson Education, Inc.

24 Chemical Kinetics First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be  k. ln [A] t =  kt + ln [A] 0 © 2012 Pearson Education, Inc.

25 Chemical Kinetics First-Order Processes This data were collected for this reaction at 198.9  C. CH 3 NCCH 3 CN © 2012 Pearson Education, Inc.

26 Chemical Kinetics First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, –The process is first-order. –k is the negative of the slope: 5.1  10  5 s  1. © 2012 Pearson Education, Inc.

27 Chemical Kinetics Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln =  kt 1/2 ln 0.5 =  kt 1/2  0.693 =  kt 1/2 = t 1/2 0.693 k Note: For a first-order process, then, the half-life does not depend on [A] 0. © 2012 Pearson Education, Inc.

28 Chemical Kinetics Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A] t = kt + 1 [A] 0 © 2012 Pearson Education, Inc.

29 Chemical Kinetics Second-Order Processes The decomposition of NO 2 at 300 °C is described by the equation NO 2 (g)NO(g) + O 2 (g) and yields data comparable to this table: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380 1212 © 2012 Pearson Education, Inc.

30 Chemical Kinetics Second-Order Processes Graphing ln vs. t, however, gives this plot Fig. 14.9(b). Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 Because this is a straight line, the process is second- order in [A]. 1 [NO 2 ] © 2012 Pearson Education, Inc.

31 Chemical Kinetics Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2  1 [A] 0 = kt 1/2 1 [A] 0 == t 1/2 1 k[A] 0 © 2012 Pearson Education, Inc.

32 Chemical Kinetics Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. © 2012 Pearson Education, Inc. Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

33 Chemical Kinetics Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. © 2012 Pearson Education, Inc.

34 Chemical Kinetics Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. © 2012 Pearson Education, Inc.

35 Chemical Kinetics Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k[NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests that the reaction occurs in two steps. NO 2 (g) + CO(g)  NO(g) + CO 2 (g) © 2012 Pearson Education, Inc.

36 Chemical Kinetics Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. © 2012 Pearson Education, Inc.

37 Chemical Kinetics Catalysts Catalysts increase the rate of a reaction. Catalysts change the mechanism by which the process occurs. They are not used up in the reaction © 2012 Pearson Education, Inc.


Download ppt "Chapter 14 Chemical Kinetics In kinetics we study the rate at which a chemical process occurs. Lecture Presentation © 2012 Pearson Education, Inc."

Similar presentations


Ads by Google